Quick Statics Question
Quick Statics Question
(OP)
I have a vehivle that I want to weigh.
I pick the front end off of the ground and get a weight from the hoist.
I do the same with the back end of the vehicle.
I measure from centerline of each axle to the pick up point and the wheel base.
Does it make sense to sum forces and moments to get the weight that is on each axle? Then adding the weight on the two axles will giv me my weight?
Thanks, this is just kind of messing with me.
I pick the front end off of the ground and get a weight from the hoist.
I do the same with the back end of the vehicle.
I measure from centerline of each axle to the pick up point and the wheel base.
Does it make sense to sum forces and moments to get the weight that is on each axle? Then adding the weight on the two axles will giv me my weight?
Thanks, this is just kind of messing with me.





RE: Quick Statics Question
Pf*L = W*(L-w) and
Pr*L = W*w.
so w = Pr*L/W
and Pf*L = W*(L-Pr*L/W)
Pf*L = WL-Pr*L
gosh ... Pf+Pr = W (possibly should've expected that, but I didn't)
another day in paradise, or is paradise one day closer ?
RE: Quick Statics Question
I am picking up the front, 43 inches away from the front axle. Back end stays on the ground. This measures 1637 lbs
Then I pick up the rear, 18 inches from the rear axle. Front stays on the ground. This measures 2728 lbs
The wheelbase of the axles is 114 inches (centerline to centerline). The length from pick up point to pick up point is 114 + 18 + 43 = 175
What is the total weight of the car?
RE: Quick Statics Question
Ted
RE: Quick Statics Question
but now you're explained the problem better, L in the two equations is not the same. I read you were lifting at the axles, but now you're not. So CG w behind the front axle; then
Pf*(43+114) = W*(114-w) and
Pr*(18+114) = W*w
so w = Pr*(18+114)/W
and Pf*(43+114) = W*(114-(Pr*(18+114)/W) )
Pf*(43+114)/114 + Pr*(18+114)/114 = W
1637*1.3772+2728*1.1579 = 5413 lbs
another day in paradise, or is paradise one day closer ?
RE: Quick Statics Question
Moments about the rear axle.
Pf x 114 = 1637 x (114 + 43) so Pf = 2254.5
Moments about the front axle.
Pr x 114 = 2728 x (114 + 18) so Pr = 3158.7
Total weight = Pf + Pr = 2254.5 + 3158.7 = 5413.2
je suis charlie
RE: Quick Statics Question
RE: Quick Statics Question
The method described will only give you an under approximation of the vehicle weight. As it assumes that the weight of the trunk and front end do not matter in each equation. (The moment of the car extending 43 inches out from the axle decreases the measured weight when lifting from the other end.) This method gives the best results when the axles are at the very ends of the vehicle or are symmetrical.
Alternately you can consider the fact that the center of gravity is different when each end is lifted because the pivot points are not symmetric. These methods assumed that the CGs were the same.
Consider this: If the axles were extremely close together the force measured at each end would be almost zero. And to the extreme case using the method provided if the car balanced on one axle at the CG the sum of the lifts would be zero and your car would be weightless.
This can be seen in the case where the CG is over the axle so that w=0 and from the equations used above
w=Pf*(114+18)/W Therefore Pf must equal zero and the weight of the car, W, can be number.
RE: Quick Statics Question
the CG of the car accounts for the total weight of the car. sure you could divide it up into portions ... ahead of the fwd axle, between the axles and behind the rear axle. sure weight on the front bumper pulls the CG fwd; so what. for calculating external reactions, if I know the CG of a bunch of weights, then applying the total weight at the cg is equivalent to applying the bunch of weights; obviously if you're interested in the internal stresses you need to consider the bunch of weights. if I have a SS beam with a triangular distributed load, I can calculate the reactions from the equivalent load at the CG; and it doesn't matter if the beam has overhangs.
another day in paradise, or is paradise one day closer ?
RE: Quick Statics Question
Try changing the values of the over hangs of the center section, 114, to extremes and you will see the equations fail reality.
w = Pr*(114+18)/W and w=114-Pf*(114+43)/W
if 114 goes to zero wouldn't the axles line up with the CG? Then the forces needed at each end to lift would be zero and the sum of the forces would be zero so your car would weigh zero???
And if it goes large
Pr*(large number+18)/W = (large number)-Pf*(Large number + 43)/W
With (large number) >>>18 or 43
Becomes Pr * Large number/W = Large number -Pf(large number)/W
Large number * (Pr+Pf)/W = Large number
(Pr+Pf)/W = 1
(Pr+Pf)= W
Funny that this is only true when the distance between the axles makes the front and rear over hangs insignificant.
RE: Quick Statics Question
if 114 goes large then the effect of the overhangs reduces, so what.
my initial post assumed lifting at the axles (no overhang) and got Pf+Pr = W (what you get from your equations if 18 and 43 go to zero); as you've derived too.
solve a SS beam, without overhangs, for a triangular load and an equivalent point load (same total load at the triangle's CG); you should get the same results. now a SS beam with overhangs and the same two loadings ... again you should get the same results (though obviously different from the preceding case).
another day in paradise, or is paradise one day closer ?
RE: Quick Statics Question
RE: Quick Statics Question
I agree with Dougt115 the method will only give an estimate.
If you lift one end and take the load reading, then as he stated that load will be reduced by the over hang at the opposite axle, however we base our calculation on the reading we have just taken and assume that that reading applies when we put the end back onto the floor, we then use it to calculate the vertical reaction at the first axle, this is not true because the overhang now adds some weight to the vertical reaction that were trying to calculate whereas when we lifted the end of the truck that overhang weight assisted.
I would post some diagrams but I'm not in a position to do that till the end of the week.
RE: Quick Statics Question
Cheers,
RE: Quick Statics Question
Thanks for your efforts, I got the same result myself and was pondering over it for a couple of days before posting, the problem isn't the centre of gravity it's the assumption that after you measure the load you assume the measured value stays the same after the truck is put down however that load would slightly increase at the axle once you put the truck down because the overhang mass or weight over the axle your pivoting about no longer assists you.
at first I agreed that it was 5413lbs but when I thought it through I agree with Doug but I still think the method gives a reasonable estimate.
RE: Quick Statics Question
another day in paradise, or is paradise one day closer ?
RE: Quick Statics Question
The car weighs 5413.2 lbs.
RE: Quick Statics Question
Thanks!
RE: Quick Statics Question
a) The CG, the lifting point and the rear axle centreline are co-linear in side view. or
b) The lifting process does not rotate the vehicle significantly
je suis charlie
RE: Quick Statics Question
On the other hand, since you are only trying to get the difference for the new rear end, lift the rear off the ground before and after the rear end is replaced. Lift at the wheels, not on the bumper. It does not matter what overhang there is or where the CG is. The difference in weights will be the rear end.